/* Problem 4--Invertible Years
   This was a very straightforward problem.  Just routinely count
   through the years until an invertible one is found. */

#include <stdio.h>
#include <stdlib.h>

int main (int argc, char **argv);
int Invert (int year);

int main (int argc, char **argv) {

  FILE *in, *out;
  int cs, year, i;

 in = fopen ("prob4.in","r");
 out = fopen ("prob4.out","w");
 cs = 0;
 while (1) { /* Read in year */
  fscanf (in,"%d",&year);
  if (year==0) break; /* Invert subsequent years one at a time. */
  for (i=year+1;!Invert(i);i++);
  fprintf (out,"Case %d: The first invertible year after %d is %d.\n",
               ++cs,year,i);
 }
 fclose (in);
 fclose (out);
 return EXIT_SUCCESS;
}

/* Inverts returns whether the specified year is invertible. */
int Invert (int year) {

  int d, inv, y;

 inv = 0; y = year;
 while (y > 0) { /* Pull digits one at a time from right */
  d = y%10;
  switch (d) { /* Build inverted number */
   case 0: inv = 10*inv;break;
   case 1: inv = 10*inv+1;break;
   case 6: inv = 10*inv+9;break;
   case 8: inv = 10*inv+8;break;
   case 9: inv = 10*inv+6;break;
   default: return 0; /* bad digit */
  }
  y /= 10;
 }
 return year == inv; /* Are they equal? */
}